Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

class Main{
	static int[][] dp;




	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);			//文字の入力
		int N = sc.nextInt();
		double[] p = new double[N];
		for(int i = 0;i < N;i++){
			p[i] = sc.nextDouble();
		}
		double[][] dp = new double[N+1][N+1];			//dp[i][j] = i番目までj枚面がでる確率
	dp[0][0] = 1;
		for(int i= 0;i < N;i++){
			for(int j = 0;j < N;j++){
				dp[i+1][j+1] += p[i] * dp[i][j];
				dp[i+1][j]   += (1-p[i])*dp[i][j];
			}
			
		}
		double ans = 0;
		for(int i = (N+1)/2;i <= N;i++){
			ans += dp[N][i];
		}
		System.out.println(ans);
		
		
	}

}







class Pair implements Comparable{
	int from;		//p
	int end;		//y
	int num;
	int bango;
	@Override
	public int compareTo(Object other) {
		Pair otherpair = (Pair)other;

		return end - otherpair.end;
	}
}